The cell potential for the given cell at 298 K
Pt| H2 (g, 1 bar) | H+ (aq) || Cu2+ (aq) | Cu(s)
is 0.31 V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu2+ is 10$-$x M. The value of x is ___________.
(Given : $E_{C{u^{2 + }}/Cu}^\Theta$ = 0.34 V and ${{2.303\,RT} \over F}$ = 0.06 V)
Answer (integer)
7
Solution
$\mathrm{Q}=\frac{\left[\mathrm{H}^{+}\right]^{2}}{\left[\mathrm{Cu}^{+2}\right] \mathrm{pH}_{2}}=\frac{10^{-6}}{\mathrm{C}} \quad \mathrm{pH}_{2}=1$
<br/><br/>
$$
\begin{aligned}
&E=E_{\text {cell }}^{\circ}-\frac{0.06}{n} \log Q \\\\
&0.31=0.34-\frac{0.06}{2} \log \frac{10^{-6}}{C} \\\\
&\log \frac{10^{-6}}{C}=1 \\\\
&C=10^{-7} M \\\\
&x=7
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
This question is part of PrepWiser's free JEE Main question bank. 127 more solved questions on Electrochemistry are available — start with the harder ones if your accuracy is >70%.